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In the linear algebra section of the Qiskit textbook appears the following claim regarding the Bloch sphere:

The surface of this sphere, along with the inner product between qubit state vectors, is a valid Hilbert space.

It is pretty clear that by scaling any quantum statevector we can easily get a vector that points outside or inside the surface of the sphere.. I.e the surface of the sphere isn't a vector space.

From where is this contradiction coming from?

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The surface of a Bloch sphere is not a Hilbert space.

Maybe they meant to write that it's a valid projective Hilbert space (in particular it's isomorphic to $\mathbb{CP}^1$)? It's not a vector space, so it cannot be a Hilbert space (note that a "projective Hilbert space" is, perhaps somewhat confusingly, not a Hilbert space).

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  • $\begingroup$ Really interesting! So what is the right way to describe the Hilbert space that single qubit states live in? $\endgroup$
    – Rajiv Krishnakumar
    Sep 28, 2022 at 20:24
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    $\begingroup$ @RajivKrishnakumar the smallest Hilbert space containing all qubits is $\mathbb{C}^2$. The point is if you want a space where the elements are in bijection (1-to-1) with qubits, then that space is not the Hilbert space $\mathbb{C}^2$, but rather the projective space $\mathbb{CP}^1$, which as glS points out, is not a Hilbert space. I don't think one is more right than the other. $\endgroup$
    – Condo
    Sep 28, 2022 at 22:02
  • $\begingroup$ Oh I see, so if I look at it in the opposite direction, is it correct to say that in addition to qubits, $\mathbb{C}^2$ also includes vectors that are not normalized and/or contain global phases? $\endgroup$
    – Rajiv Krishnakumar
    Sep 29, 2022 at 10:04
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    $\begingroup$ @RajivKrishnakumar the projective space $\mathbb{CP}^1$ is what you get from $\mathbb{C}^2$ modulo an equivalence relation. In other words, if you take $\mathbb{C}^2$, and identify all points which only differ by a global phase and a normalisation factor (and define a suitable topology), you get $\mathbb{CP}^1$. So in this sense, yes, $\mathbb C^2$ includes vectors that are not normalizes and "contain global phases" (the latter does'nt actually mean much; you should say that it contains different vectors differering by a global phase, which all correspond to the same physical state) $\endgroup$
    – glS ♦
    Sep 29, 2022 at 10:19
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    $\begingroup$ Got it, makes sense, thank you for taking the time everyone for the clear explanations! I gave quick shout to the qiskit folks pointing them to this thread. They may already know this point and decided to "simplify" things when teaching it, but either way, I though it wouldn't hurt to let them know :) $\endgroup$
    – Rajiv Krishnakumar
    Sep 29, 2022 at 10:36

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